如何为可重用的 Django 应用程序进行迁移?

问题描述

我正在制作一个没有项目的可重用 Django 应用程序.这是目录结构:

I am making a reusable Django app without a project. This is the directory structure:

/
/myapp/
/myapp/models.py
/myapp/migrations/
/myapp/migrations/__init__.py

当我运行 django-admin makemigrations 时出现以下错误:

When I run django-admin makemigrations I get the following error:

django.core.exceptions.ImproperlyConfigured: Requested setting DEFAULT_INDEX_TABLESPACE, but settings are not configured. You must either define the environment variable DJANGO_SETTINGS_MODULE or call settings.configure() before accessing settings.

显然，这是因为我没有配置设置模块，因为这是一个可重复使用的应用程序.但是，我仍然想使用我的应用程序进行迁移.我怎样才能制作它们?

Obviously, this is because I don't have a settings module configured, because this is a reusable app. However, I would still like to ship migrations with my app. How can I make them?

推荐答案

其实你不需要有项目，你只需要设置文件和脚本，运行迁移创建.设置必须包含以下内容(最少):

Actually, you don't need to have project, all you need is settings file and script, that runs migrations creation. Settings must contain folowing (minimum):

# test_settings.py

DEBUG = True

SECRET_KEY = 'fake-key'

INSTALLED_APPS = [
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'your_app'
]

进行迁移的脚本应该如下所示:

And the script, that makes migrations should look like this:

# make_migrations.py

import os
import sys

if __name__ == "__main__":
    os.environ.setdefault("DJANGO_SETTINGS_MODULE", "test_settings")
    from django.core.management import execute_from_command_line
    args = sys.argv + ["makemigrations", "your_app"]
    execute_from_command_line(args)

你应该通过 python make_migrations.py 运行它.希望它可以帮助某人！

and you should run it by python make_migrations.py. Hope it helps someone!

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